In this HCF and LCM worksheets 2,we give the questions on HCF and LCM whose qualities are in high standard and practicing these questions will definitely make the students to score marks in HCF and LCM.
Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 |
Since the H.C.F is 13, the two numbers could be 13x and 13y
Their product is 2028. So, (13x)X(13y) = 2028 ===> xy = 12 We have to find the values of "x" and "y" such that their product is 12. The possibles values of (x,y) = (1,12),(2,6),(3,4). Here, we have to check an important thing. That is, in the above pairs of values of (x,y), which are all co-primes? [Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1] Therefore in the above pairs, (1,12) and (3,4) are the co-primes. Hence, the number of pairs is 2
Since the H.C.F is 49, the two numbers could be 49x and 49y
Their sum is 588. So, 49x+49y = 588 ===> x+y = 12 We have to find the values of "x" and "y" such that their sum is 12. The possibles values of (x,y) = (1,11),(2,10),(3,9),(4,8)(5,7)(6,6). Here, we have to check an important thing. That is, in the above pairs of values of (x,y), which are all co-primes? [Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1] Therefore in the above pairs, (1,11) and (5,7) are the co-primes. Hence, the number of pairs is 2
For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.
To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465.That is nothing but the H.C.F of (403,434,465) The H.C.F of (403,434,465) = 31 liters Each cask must be of the volume 31 liters. Req. No. of casks = (403/31)+(434/31)+(465/31)= 13+14+15 = 42 Hence, the least possible number of casks of equal size required is 42
We require the least number of square tiles. So, each tile must be of maximum dimension.
To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32.That is nothing but the H.C.F of (16.58, 8.32) 16.58X100 = 1658 cm and 8.32X100 = 832 cm H.C.F of (1658, 832) = 2 Hence the side of the square tile is 2 cm Required no. of tiles = (Area of the floor) / (Area of a square tile) = (1658X832)/2X2 = 344864 Hence, the least number of square tiles required = 344864
To answer this question, we have to find the least number which is exactly divisible by the given numbers 15,20 and 25.That is nothing but the L.C.M of (15,20,25)
L.C.M of (15,20,25) = 300 So, we need 300 soldiers such that they stand in rows of 15, 20 , 25. But, it has to form a perfect square (as per the question) To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square. To make 300 as perfect square, we have to multiply 300 by 3. Then, it is 900 which is a perfect square. Hence, the least number of soldiers required is 900.
The given two quantities 120 and 180 can be divided by 10, 20,... exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,... ltrs.
But, the target of the question is, the volume of oil filled in tins must be greatest. So, we have to find the largest number which exactly divides 120 and 180.That is nothing but the H.C.F of (120, 180) H.C.F of (120, 180) = 60 The 1st kind 120 ltrs is sold in 2 tins of of volume 60 ltrs in each tin. The 2nd kind 180 ltrs is sold in 3 tins of of volume 60 ltrs in each tin. Hence, the greatest volume of each tin is 60 ltrs.
For example, let the two signals change after every 3 secs and 4 secs respectively.
Then the first signal changes after every 3, 6, 9, 12 seconds... Like this, the second signal changes after every 4, 8, 12 seconds... So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the L.C.M of (48,72,108). L.C.M of (48,72,108) is 432 seconds = 7 min 12 sec So, after every 7 min 12 sec, all the signals will change simultaneously. At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs. Hence, three signals will change simultaneously at 8:27:12 seconds.
For example, let the two bells toll after every 3 secs and 4 secs respectively.
Then the first bell tolls after every 3, 6, 9, 12 seconds... Like this, the second bell tolls after every 4, 8, 12 seconds... So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds The same thing happened in our problem. To find the time, when they will all toll together, we have to find the L.C.M of (2,4,8,6,10,12). L.C.M of (2,4,8,6,10,12) is 120 seconds = 2 minutes. So, after every two minutes, all the bell will toll together. For example, in 10 minutes, they toll together 10/2 = 5 times That is, after 2,4,6,8,10 minutes. It does not include the one at the start Similarly, in 30 minutes, they toll together 30/2 = 15 times.(excluding one at the start)
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Therefore, required number (9999 - 399) = 9600.
Given: Three numbers are in the ratio 3:4:5.
Then the numbers are 3x, 4x, 5x L.C.M of (3x, 4x, 5x) = 60x ---(1) Given L.C.M = 2400 ---(2) From (1)&(2), 60x = 2400 ---> x = 40 So, the number are (3x40, 4x40, 5x40) Hence, the required H.C.F is 40 |